At $11$ : $55\text{ p.m.}$, Thomas ties a weight to the minute hand of a clock. The clockwise torque applied by the weight (i.e. the force it applies on the clock's hand to move clockwise) varies in a periodic way that can be modeled by a trigonometric function. The torque peaks $15$ minutes after each whole hour, when the minute hand is pointing directly to the right, at $3\text{ Nm}$ (Newton metre, the SI unit for torque). The minimum torque of $-3\text{ Nm}$ occurs $15$ minutes before each whole hour, when the minute hand is pointing directly to the left. Find the formula of the trigonometric function that models the torque $\tau$ applied by the weight $t$ minutes after Thomas attached the weight. Define the function using radians. $ \tau(t) = $
Solution: Let's start by writing a formula for the torque applied by the weight $u$ minutes after midnight. Both sine and cosine can be used to model periodic contexts. We can decide which is better fitting by considering the $y$ -intercept. The sine function intercepts the $y$ -axis at its midline, and the cosine function intercepts the $y$ -axis at its peak. The clockwise torque applied by the weight is halfway from its minimum to its maximum at $u = 0$ minutes after midnight, so we'll use a sine function. The torque applied by the weight has period $60$ minutes, since every hour it reaches its maximum and minimum once. The midline of the torque is halfway between its maximum value and its minimum value, or $\dfrac{3 + (-3)}{2} =0$ Its amplitude is half the difference between its maximum and minimum values, or $\dfrac{3 - (-3)}{2} =3$ Since the ordinary sine function $f(u) = \sin u$ has period $2\pi$, midline $y = 0$, and amplitude $1$, we can stretch it horizontally by a factor of ${\dfrac{60}{2\pi}}$, and stretch it vertically by a factor of ${3}$. We don't need to shift it up or down since its midline is already $ y = 0$. $ \tau(u) = {3}\sin\left({\dfrac{2\pi}{60}}u\right)$ Since the midnight is $5$ minutes after Thomas attached the weight, a time that's $t$ minutes after Thomas attached the weight is $t - 5$ minutes after midnight. $ \tau(t) = {3}\sin\left({\dfrac{2\pi}{60}}(t - 5)\right)$ The function $ \tau(t) = {3}\sin\left({\dfrac{2\pi}{60}}(t-5)\right)$ has period $60$, amplitude $3$, and midline $y = 0$, and it is halfway from its minimum to its maximum at time $t=5$. So it's a good estimate for the torque applied by the weight $t$ minutes after midnight.